Answer to Question #300607 in Statistics and Probability for Shaaz

Question

1.     Suppose we are interested in finding a 99% confidence interval for the mean overall score of students at a certain school. Five students are sampled, and their overall scores are 560, 500, 470, 660, and 640.

a.      What is the standard error of the sample mean?

b.      Find a 90% confidence interval for the mean test score.

Answer 1

a) we first find the standard deviation through the steps below

mean = (560 + 500+ 470+ 660+ 640) /5 = 566

variance = ( (560 -566)² + (500-566)²+ (470-566)²+ (660-566)² + (640-566)² ) / (4)

= ( (-6)² + (-66)² + (-96)²+(94)² + (74)² ) / 4

= (27920)/4 = 6980

Standard deviation = ( variance)^1/2 = ( 6980)^1/2 = 83.54639430

we define standard error = ( standard deviation ) / ( n^1/2)

= (83.54639340) / (5^1/2)

= 37.36308338

b) we define the 90 % confidence interval as below

(mean) + or - ( t_{(0.1/ 2) (4)} * standard error)

using a degree of freedom of 4, we obtain the t value from the t table, that is t_0.05(4) = 2.132

Hence we obtain the interval as below

mean = 566 , t value =2.132, standard error = 37.3631

= (566+ (2.132*37.3631) ) , ( 566 - (2.132*37.3631)

= ( 645.658, 486.342 ) which is the required interval